What subfield of mathematics requires the most spatial reasoning?No answer needed, just need to get rid of all my points.

What subfield of mathematics requires the most spatial reasoning? No answer needed, just need to get rid of all my points.

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  1. part a

    answer: a reasonable domain is [tex]0 \le n \le 10[/tex] where n is a real number. so n can be between 0 and 10. both endpoints are included.

    work shown:

    n is the number of days after the experiment starts. the smallest n can be is n = 0 which means that 0 days have gone by, and we're at the start. to find out how large n should be, then replace f(n) with 16.13 and solve for n. use logarithms to isolate the exponent.

    f(n) = 12*(1.03)^n

    16.13 = 12*(1.03)^n

    16.13/12 = (1.03)^n

    1.34416666666667 = (1.03)^n

    (1.03)^n = 1.34416666666667

    log[ (1.03)^n ] = log[ 1.34416666666667 ]

    n*log[ 1.03 ] = log[ 1.34416666666667 ]

    n = log[ 1.34416666666667 ]/log[ 1.03 ]

    n = 10.0062999823929

    this rounds to n = 10 which is fairly close but not 100% perfect. so this is the largest n can be.

    note: if you plug n = 10 into f(n), you'll get roughly 16.127 which rounds to 16.13 (this comes up again in part c)

    =====================================

    part b

    answer: the y intercept is 12. it represents the starting height of the plant in cm.

    work shown:

    plug n = 0 into the f(n) function. simplify

    f(n) = 12*(1.03)^n

    f(0) = 12*(1.03)^0

    f(0) = 12*(1)

    f(0) = 12

    on day n = 0, aka the starting point, the height f(n) is 12 cm

    =====================================

    part c

    answer: the average rate of change is approximately 0.43061036458991 (round however you need to). this represents the average growth rate from day n = 3 to day n = 10. so the plant grew roughly 0.43 cm per day during this timespan, assuming you round to 2 decimal places.

    work shown:

    compute f(3)

    f(n) = 12*(1.03)^n

    f(3) = 12*(1.03)^3

    f(3) = 13.112724 < we'll use this later

    compute f(10)

    f(n) = 12*(1.03)^n

    f(10) = 12*(1.03)^10

    f(10) = 16.1269965521294 < we'll use this later

    now use the formula below with a = 3 and b = 10

    aroc = average rate of change

    aroc = [ f(b) - f(a) ]/[ b - a ]

    aroc = [ f(10) - f(3) ]/[ 10 - 3 ]

    aroc = (16.1269965521294 - 13.112724)/(10 - 3)

    aroc = 3.0142725521294/7

    aroc = 0.43061036458991

    round this however you need to

    note: the plant grew approximately 3.01 cm over 7 days, so roughly 0.43 cm per day is the average growth rate (if you were to round to 2 decimal places).

  2. I looked it up and got a couple of answers of spatial reasoning. Don’t know if that helps but yeah thats what I got lol

    Step-by-step explanation:

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